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Anestetic [448]
3 years ago
9

What is the mass of 564 liters of oxygen in kilograms?

Chemistry
1 answer:
Alika [10]3 years ago
6 0
One kg of oxygen is 0.3977 liters so:

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The concentration of dye in Solution A is 25.527 M. You have 13 mL of water at your disposal to make the dilutions. The solution
Lady bird [3.3K]

Answer:

 38825.472 M

Explanation:

Using

C1V1=C2V2

To make solution B,

C1 = 25.527 M, V1 = 13 mL

C2 = 25.527 x 8 = 204.216 M

V2 = 13 X 12 = 156 mL

To make solution C'

C1 = 8 X 202.216 M = 1617.728

V1 = 8 X 156mL = 1248 mL

V2 = 4 X 13 = 52 mL

C2 = C1V1 / V2

    =1617.728 X 1248 / 52

 =38825.472 M

7 0
4 years ago
Calculate the moles of 336 L of carbon dioxide gas (CO2) at STP.
fenix001 [56]

Answer:

14.99 mols of CO2

Explanation:

STP= 273 k , 1 atm , .0821 gas constant

N=PV/RT

N= 1x336/.0821x273

14.99 mols of CO2

659.6 grams of CO2

8 0
3 years ago
HOW DO THE FOLLOWING COMPARE IN THE AMOUNT OF ALCOHOL THEY CONTAIN: 12 OZ. BEER(5% ALCOHOL), 12 OZ. WINE COOLER(5% ALCOHOL), 1 1
Pavlova-9 [17]

The amount of alcohol contained in each of the listed beverages can be compared as follows; 1/2 oz of 80 proof liquor > 5 oz of wine (12% alcohol) > 12 oz of wine cooler (5%alcohol) = 12 oz of beer  (5%alcohol) .

A drink that contains alcohol must be an intoxicating beverage. The extent of intoxication of an alcoholic beverage depends on the amount of alcohol that the beverage contains. There are various types of alcoholic beverages as listed in the question.

The amount of alcohol contained in each of the listed beverages can be compared as follows; 1/2 oz of 80 proof liquor > 5 oz of wine (12% alcohol) > 12 oz of wine cooler (5%alcohol) = 12 oz of beer  (5%alcohol) .

Learn more about alcoholic beverage: brainly.com/question/6967136

8 0
3 years ago
How many moles of Ca[OH]2 are present in 80.0 g of Ca[OH]2?​
artcher [175]

Answer:

1.08mol

Explanation:

moles = reacting mass/ molecular weight

reacting mass = 80.0g molecular weight of Ca[OH]2= 40 + 2(16 +1) = 74g/mol

mole = 80.0/74 = 1.08mol

4 0
4 years ago
A gas occupies 525 mL at a pressure of 45.0 kPa. What would the volume of the gas be at a pressure of 65.0 kPa
choli [55]

Answer:

The volume of the gas at a pressure of 65.0 kPa would be 363 mL

Explanation:

Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:

P*V=k

Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:

P1*V1=P2*V2

In this case:

  • P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
  • V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
  • P2= 65 kPa= 65,000 Pa
  • V2= ?

Replacing:

45,000 Pa* 0.525 L= 65,000 Pa*V2

Solving:

V2=\frac{45,000 Pa* 0.525 L}{65,000 Pa}

V2=0.363 L=363 mL

<u><em>The volume of the gas at a pressure of 65.0 kPa would be 363 mL</em></u>

6 0
3 years ago
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